# Rolle's Theorem Questions and Examples

Rolle's theorem is a special case of the mean value theorem. It is discussed here through examples and questions.

## Rolle's Theorem

Rolle's theorem is the result of the mean value theorem where under the conditions:\( f(x) \) be a continuous functions on the interval \( [a, b] \) and differentiable on the open interval \( (a, b) \) , there exists at least one value c of x such that \[ f '(c) = \dfrac{f(b) - f(a)}{b - a} \] Now if the condition \( f(a) = f(b) \) is satisfied, then the above simplifies to : \( f'(c) = 0 \).

In other words, under the above three conditions we can always find a tangent to the curve of f that is horizontal (slope = \( f'(c) = 0 \) ).

Rolle's Theorem

If \( f(x) \) is

1) a continuous function on the interval \( [a, b] \)

2) differentiable on the open interval \( (a, b) \)

3) and \( f(a) = f(b) \),

then there is at least one value c of x in the interval \( (a, b) \) such that

\( f'(c) = 0 \)

1) a continuous function on the interval \( [a, b] \)

2) differentiable on the open interval \( (a, b) \)

3) and \( f(a) = f(b) \),

then there is at least one value c of x in the interval \( (a, b) \) such that

\( f'(c) = 0 \)

## Example 1

The graph of \( f(x) = - x^2 + 6x - 6 \) for \( 1 \leq x \leq 5 \) is shown below. \( f(1) = f(5) = - 1 \) and \( f \) is continuous on [1 , 5] and differentiable on (1 , 5) hence, according to Rolle's theorem, there exists at least one value of \( x = c \) such that \( f '(c) = 0 \).\( f '(x) = - 2 x + 6 \)

\( f '(c) = - 2 c + 6 = 0 \)

Solve the above equation to obtain

\( c = 3 \)

Therefore at \( x = 3 \) there is a tangent to the graph of \( f \) that has a slope equal to zero (horizontal line) as shown in figure 1 below.

## Example 2

The graph of \( \quad f(x) = \sin(x) + 2 \) for \( \quad 0 \leq x \leq 2\pi \) is shown below. \( \quad f(0) = f(2\pi) = 2 \) and \( f \) is continuous on [0 , \( 2\pi \)] and differentiable on (0 , \( 2\pi \)) hence, according to Rolle's theorem, there exists at least one value ( there may be more than one! ) of \( x = c \) such that \( f '(c) = 0 \).\( f '(x) = \cos(x) \)

\( f '(c) = \cos(c) = 0 \)

The above equation has two solutions on the interval [0 , \( 2\pi \)]

\( c_1 = \dfrac{\pi}{2} \) and \( c_2 = \dfrac{3\pi}{2} \).

Therefore both at \( x = \dfrac{\pi}{2} \) and \( x = \dfrac{3\pi}{2} \) there are tangents to the graph that have a slope equal to zero (horizontal line) as shown in figure 2 below.

## Example 3

Function \( f \) in figure 3 does not satisfy Rolle's theorem: although it is continuous and \( f(-1) = f(3) \), the function is not differentiable at \( x = 1 \) and therefore \( f '(c) = 0 \) with \( c \) in the interval (-1 , 3) is not guaranteed. In fact it is easy to see that there is no horizontal tangent to the graph of \( f \) on the interval (-1 , 3).## Question 1

Which of the functions given below satisfy all three conditions of Rolle's theorem?a) \( f(x) = \cos(x) \), for \( x \) in \( [0 , 2\pi \)] \)

b) \( g(x) = |x - 2| \), for \( x \) in \( [0 , 4] \)

c) \( h(x) = 1 / x^2 \), for \( x \) in \( [-1 , 1] \)

d) \( k(x) = |\sin(x)| \), for \( x \) in \( [0 , 2\pi ] \)

### solution to question 1

a)\( f(0) = 1 \) and \( f(2\pi) = 1 \) therefore \( f(0) = f(2\pi) \)

\( f \) is continuous on \( [0 , 2\pi ] \)

Function \( f \) is differentiable in \( (0 , 2\pi ) \)

Function \( f \) satisfies all conditions of Rolle's theorem

b)

function \( g \) has a V-shaped graph with vertex at \( x = 2 \) and is therefore not differentiable at \( x = 2 \).

Function \( g \) does not satisfy all conditions of Rolle's theorem

c)

Function \( h \) is undefined at \( x = 0 \).

Function \( h \) does not satisfy all conditions of Rolle's theorem.

d)

\( k(x) = |\sin(x)| \), for \( x \) in \( [0 , 2\pi] \)

The graph of function \( k \) is shown below and it shows that function \( k \) is not differentiable at \( x = \pi \).

Function \( k \) does not satisfy all conditions of Rolle's theorem.

## Question 2

Check that function \( f(x) = x^2 - 4x + 3 \) on the interval [1 , 3] satisfies all conditions of Rolle's theorem and then find all values of \( x = c \) such that \( f '(c) = 0 \).### solution to question 2

\( f \) is a polynomial function and is therefore continuous on the interval [1 , 3] and differentiable on the interval (1 , 3). Also \( f(1) = f(3) = 0 \) and therefore function \( f \) satisfies all three conditions of Rolle's theorem and there is at least one value of \( x = c \) such that \( f '(c) = 0 \).\( f '(x) = 2x - 4 \)

\( f '(c) = 2c - 4 = 0 \)

Solve for \( c \) to obtain

\( c = 2 \).

The graph below shows \( f \), \( f' \) and the tangent at \( x = c = 2 \) is horizontal; and \( f'(2) = 0 \)

## Question 3

Check that function \( g(x) = \cos(x) \) on the interval \( [- \pi /2 , 3\pi/2] \) satisfies all conditions of Rolle's theorem and then find all values \( x = c \) such that \( g '(c) = 0 \).### solution to question 3

Function \( g \) is a cosine function and is therefore continuous on the interval \( [- \pi/2 , 3\pi/2] \) and differentiable on the interval \( (- \pi /2 , 3\pi /2) \) . Also \[ g(- \pi /2) = g( 3\pi /2) = 0 \] and therefore function \( g \) satisfies all three conditions of Rolle's theorem and there is at least one value of \( x = c \) such that \( f '(c) = 0 \).\( g '(x) = - \sin(x) \)

\( g '(c) = - \sin(c) = 0 \)

Solve for \( c \) to obtain

\( c = n \pi \) , \( n = 0,\pm 1 , \pm 2 , ... \)

Solutions on the interval \( [- \pi/2 , 3\pi /2] \) are

\( c_1 = 0 \) and \( c_2 = \pi \);

The graph below shows \( g \), \( g' \) and the tangents at \( x = c_1 = 0 \) and \( x = c_2 = \pi \) are horizontal; and \( g'(0) = 0 \) and \( g'(\pi) = 0 \)

__More References and links__

problems related to the mean value theorem Continuous Functions in Calculus

Non Differentiable Functions

calculus